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\title[ch05]{Chapter 02: IDEAL STRUCTURE OF THE WEYL ALGEBRA }
\author[]{SCC}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% 封面页
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% 目录页
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% Section 1
\section{THE DEGREE OF AN OPERATOR.}

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\begin{frame}{1.0 DEFINITION. }

Let $D \in A_n$. The {\color{red}degree} of $D$ is the largest length of the multi-indices $(\alpha,\beta) \in \mathbb{N}^n \times \mathbb{N}^n$ for which $x^\alpha \partial^\beta$ appears with non-zero coefficient in the canonical form of $D$. It is denoted by $\deg(D)$. 

As with the degree of a polynomial, we use the convention that the zero polynomial has degree $-\infty$. An example will suffice: the degree of $2x_1 \partial_2 + x_1 x_2^3 \partial_1 \partial_2$ is 6.

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\begin{frame}{1.1 THEOREM. }

The degree satisfies the following properties; for $D,D{\,}' \in A_n$:
\begin{enumerate}
    \item $\deg(D+D{\,}') \leq \max\{\deg(D),\deg(D{\,}')\}$.
    \item $\deg(DD{\,}') = \deg(D) + \deg(D{\,}')$.
    \item $\deg([D,D{\,}']) \leq \deg(D) + \deg(D{\,}') - 2$.
\end{enumerate}

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\begin{frame}{1.2 COROLLARY. }

The algebra $A_n$ is a domain.

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% Section 2
\section{IDEAL STRUCTURE.}
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\begin{frame}{2.1 THEOREM. }

The algebra $A_n$ is simple, i.e., the only proper two-sided ideal is zero. 

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\begin{frame}{2.2 COROLLARY. }

Every endomorphism of $A_n$ is injective.

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\begin{frame}{2.3 THEOREM BY STAFFORD. }

The Weyl algebra is not a left principal ideal ring. 

For example, the left ideal generated by $\partial_1, \partial_2$ in $A_2$ is not principal. 

However, every left ideal of $A_n$ is generated by two elements. 

This is a very important result, due to J.T. Stafford. 


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% Section 3
\section{POSITIVE CHARACTERISTIC.}
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\begin{frame}{3.1 EXAMPLE. }

Consider first the algebra of operators $R_1$ generated by $\mathbb{Z}_p[x]$ and its derivation $\partial$. Let us calculate $\partial^p(x^k)$. If $k < p$, then $\partial^p(x^k) = 0$. If $k \geq p$, then
\[
\partial^p(x^k) = k \ldots (k - p + 1)x^{k-p} = 0
\]

since the coefficient is divisible by $p$. Thus $\partial^p = 0$ as an operator on $\mathbb{Z}_p[x]$. We conclude that the ring of operators $R_1$ has nilpotent elements. In particular it is not a domain.

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\begin{frame}{3.2 EXAMPLE. }

Now consider the ring $R_2$ generated over $\mathbb{Z}_p$ by $z_1, z_2$ subject to the relation $[z_2,z_1] = 1$. This is a domain. However, it is not like the Weyl algebra in characteristic zero in another respect: it is not a simple ring. For example, if $f \in \mathbb{Z}_p[z_1]$ then $[z_2,f] = \partial f / \partial z_1$. In particular,

\[
[z_2,z_1{}^p] = pz_1{}^{p-1} = 0
\]

over $\mathbb{Z}_p$. It follows that $z_1{}^p$ commutes with every element of $R_2$. Hence the left ideal generated by $z_1{}^p$ in $R_2$ is a two-sided ideal. In particular $R_2$ is not a simple ring.

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% Section 4
\section{EXERCISES.}
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\begin{frame}{4.1 EXERCISE. }

Let $n \geq 2$. Show that the left ideal $L$ of $A_n$ generated by $\partial_1, \ldots, \partial_n$ is not principal.

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Hint: Show that a generator would have degree 1, and obtain a contradiction from that.

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\begin{frame}{4.2 EXERCISE. }

Show that the left ideal of $A_3$ generated by $\partial_1, \partial_2, \partial_3$ may also be generated by two elements.

Hint: Choose $D_1 = \partial_1$ and $D_2 = \partial_2 + x_1 \partial_3$ to be the generators, and calculate $[D_1, D_2]$.

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\begin{frame}{4.3 EXERCISE. }

Let $K(X)$ be the rational function field in $n$ variables. Let $B_n(K)$ be the $K$-subalgebra of $End_K(K(X))$ generated by the elements of $K(X)$ and the derivations $\partial_1, \ldots, \partial_n$. 

An element $d \in B_n(K)$ may be written in the form $d = \sum_\alpha q_\alpha \partial^\alpha$, where $q_\alpha \in K(X)$. 

The {\color{red}order} of $d$, denoted by $ord(d)$, is the largest $|\alpha|$ for which $q_\alpha \neq 0$. 

Show that if $d{\,}' \in B_n(K)$ then:
\begin{enumerate}
    \item $ord(d + d{\,}') \leq \max\{ord(d), ord(d{\,}')\}$.
    \item $ord(dd{\,}') = ord(d) + ord(d{\,}')$.
\end{enumerate}

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\begin{frame}{4.4 EXERCISE. }

Use the order defined above to show that:
\begin{enumerate}
    \item $B_n(K)$ is a domain.
    \item $B_n(K)$ is a simple ring.
\end{enumerate}

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\begin{frame}{4.5 EXERCISE. }

Show that $B_1(K)$ admits a left division algorithm. That is, if $a,b \in B_1(K)$, then there exist $q,r \in B_1(K)$ such that $a = qb + r$ and $ord(r) < ord(b)$. Use this to prove that every left ideal of $B_1(K)$ is principal.

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\begin{frame}{4.6 EXERCISE. }

If $n \geq 2$, is every left ideal of $B_n(K)$ principal?

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\begin{frame}{4.7 EXERCISE. }

Let $A_1(\mathbb{Z})$ be the $\mathbb{Z}$-subalgebra of $A_1(\mathbb{Q})$ generated by $x$ and $\partial$. Show that $A_1(\mathbb{Z})$ is not a simple ring.

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\begin{frame}{4.8 EXERCISE. }

Let $d \in A_1(K)$, and write $d = \sum_{i=0}^{m} g_i(x) \partial^i$. 

Call $m$ the {\color{red}order} of $d$ and $L(d) = g_m(x) \neq 0$ its {\color{red}leading term}. 

If $J$ is a left ideal of $A_1(K)$, show that:
\begin{enumerate}
    \item The set $L_n(J) = \{L(d) : d \in J \text{ and } d \text{ has order } n\} \cup \{0\}$ is an ideal of $K[x]$.
    \item If $m \leq n$, then $L_m(J) \subseteq L_n(J)$.
    \item $J$ is principal if and only if $L_n(J) = L_{n+1}(J)$ for all $n > 0$ such that $L_n(J) \neq 0$.
\end{enumerate}

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\begin{frame}{4.9 EXERCISE. }

Show that the left ideal $A_1 \partial^2 + A_1 (x\partial - 1)$ is not a cyclic ideal of $A_1(K)$.

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\begin{frame}{4.10 EXERCISE. }

Let $\phi: A_1^2 \to A_1$ be the map defined by $\phi(a,b) = a\partial + bx$. 

Show that $\phi$ is surjective and that its kernel is isomorphic to the left ideal 
$$A_1 \partial^2 + A_1 (x\partial + 2). $$

Conclude that this is a projective left ideal of $A_1(K)$.

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\begin{frame}{4.11 EXERCISE. }

Let $R_2$ be the ring defined in §3. Show that:
\begin{enumerate}
    \item $R_2$ is a domain.
    \item The centre of $R_2$ is $\mathbb{Z}_p[z_1^p, z_2^p]$.
    \item $R_2$ is finitely generated as a module over its centre.
\end{enumerate}

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